Question
Question 2 Consider two charges, where one (+3.80nC) is at the origin and the other (-14.4nC) is at the position x=2.86 cm. Find the xcoordinate where a proton would experience zero net force. Answer: square disappointed
Solution
4.4
(316 Votes)
Hannah
Veteran · Tutor for 9 years
Answer
Here's how to find the x-coordinate where a proton would experience zero net force:**1. Understand the Problem**We have two charges: a positive charge (q1) at the origin and a negative charge (q2) at x = 2.86 cm. A proton (positive charge) will be repelled by the positive charge and attracted by the negative charge. We need to find the point on the x-axis where these two forces balance each other out.**2. Set up the Equation**Let the position of the proton be x. The force exerted on the proton by q1 is:F1 = k * q1 * qp / x^2where:* k is Coulomb's constant (8.99 x 10^9 N m^2/C^2)* q1 is the first charge (+3.80 nC = 3.80 x 10^-9 C)* qp is the charge of the proton (1.60 x 10^-19 C)* x is the distance between the proton and q1The force exerted on the proton by q2 is:F2 = k * q2 * qp / (2.86 cm - x)^2 = k * q2 * qp / (0.0286 m - x)^2where:* q2 is the second charge (-14.4 nC = -14.4 x 10^-9 C)* 0.0286 m - x is the distance between the proton and q2 (converted cm to meters)For the net force to be zero, F1 and F2 must be equal in magnitude:F1 = F2k * q1 * qp / x^2 = k * |q2| * qp / (0.0286 - x)^2**3. Solve for x**Notice that k and qp cancel out:q1 / x^2 = |q2| / (0.0286 - x)^2Substitute the values of q1 and q2:(3.80 x 10^-9) / x^2 = (14.4 x 10^-9) / (0.0286 - x)^2Simplify and cross-multiply:3.80 * (0.0286 - x)^2 = 14.4 * x^2Take the square root of both sides:sqrt(3.80) * (0.0286 - x) = sqrt(14.4) * x1.949 * (0.0286 - x) = 3.795 * x0.0558 - 1.949x = 3.795x0.0558 = 5.744xx = 0.0558 / 5.744x ≈ 0.00971 m = 0.971 cmSince the negative charge is larger in magnitude, the point of zero force must be to the left of the positive charge (closer to the origin). Therefore, the x-coordinate is approximately -0.971 cm.**Answer:** -0.971 cm